Question

The octane rating of gasoline is a relationship of the burning efficiency of the given gasoline...

1. The octane rating of gasoline is a relationship of the burning efficiency of the given gasoline mixture to the burning efficiency of octane (C8H18). Like most of hydrocarbons, octane reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction:

C8H18 (l) + O2(g) → CO2 (g) + H2O (g)

If 0.240 mol of octane is allowed to react with 0.890 mol of oxygen gas,

1. how many moles of water are produced in this reaction (theoretical yield of water)?
2. How much excess reactant (E.R.) is left in moles?

The combustion of octane occcurs as follows:
2C8H18 (l) + 25O2(g) → 16CO2 (g) + 18H2O (g)
It is evident from the balanced equation that:
2 mols of C8H18 require 25 mol of O2 to react completely.
1 mol of C8H18 requires 12.5 mol of O2 to react completely.
0.240 mol of C8H18 requires O2 = 12.5 x 0.240 = 3
However, only 0.89 mol of O2 is available.
Hence, O2 is the limiting reagent and C8H18 is the excess reagent.
The quantity of product formed depends on the quantity of the limiting reagent.
25 mol of O2 to react to produce H2O = 18 mol
1 mol of O2 to react to produce H2O = 18/25 mol
0.89 mol of O2 to react to produce H2O = (18/25) x 0.89 = 0.6408 mol
0.89 mol of O2 reacts with C8H18 = (2/25) x 0.89 = 0.0712 mol
Number of moles of (excess reagent) left unreacted = 0.240 -0.0712 = 0.1688 mol